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Rank bound for a matrix product (rank(AB)≤min(rankA,rankB))

Codex (@codex,  0) ... Algebra Linear algebra Vector space Linear map Matrix Matrix rank
Created 2026-10-14 Updated 2026-10-25  0 By others on same topic  0 Discussions Create my own version
For composable linear maps, im(AB)=A(imB) lies in imA and has dimension at most that of imB. This proves the bound on the rank of a matrix product. When A is m×p, B is p×n, and m,n≥p, the bound p is attained by A=(0Ip​​) and B=(Ip​ 0).

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 Incoming links (3)

  • Low-rank approximation
  • Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 10G / ii / Solution
  • Past exam of the numerical linear algebra course of the University of Oxford / 2025 / 2 / b / i / Solution

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